Univ. Beograd. Publ. Elektrotehn. Fak.
Ser. Mat. 14 (2003), 20–25.
SOME IDENTITIES
FOR THE RIEMANN ZETA-FUNCTION
Aleksandar Ivić
Several identities for the Riemann zeta-function ζ(s) are proved. For example, if s = σ + it and σ > 0, then
∞ (1 − 21−s )ζ(s) 2
dt = π (1 − 21−2σ )ζ(2σ).
s
σ
−∞
∞
Let as usual ζ(s) = n=1 n−s (e s > 1) denote the Riemann zeta-function.
The motivation for this note is the quest to evaluate explicitly integrals of |ζ( 21 +
it)|2k , k ∈ N, weighted by suitable functions. In particular, the problem is to
evaluate in closed form
∞
√
k
dt
3 − 8 cos(t log 2) |ζ( 21 + it)|2k (k ∈ N).
1
2 k
0
4 +t
When k = 1, 2 this may be done, thanks to the identities which will be established
below. The first identity in question is given by
Theorem 1. Let s = σ + it. Then for σ > 0 we have
∞
(1 − 21−s )ζ(s) 2
dt = π (1 − 21−2σ ) ζ(2σ).
(1)
s
σ
−∞
Since lims→1 (s − 1)ζ(s) = 1, then setting in (1) σ =
Corollary 1.
(2)
0
∞
3−
1
2
we obtain the following
√
dt
8 cos(t log 2) |ζ( 21 + it)|2 1
= π log 2.
+
t2
4
2000 Mathematics Subject Classification: 11M06
Keywords and Phrases: The Riemann zeta-function, identities, characteristic function.
20
21
Some identities for the Riemann zeta-function
Another identity, which relates directly the square of ζ(s) to a Mellin-type
integral, is contained in
Theorem 2. Let χA (x) denote the characteristic function of the set A, and let
∞ x
∞ x
du
(x 1).
(3)
ϕ(x) :=
χ[2m−1,2m)
χ[2n−1,2n) (u)
u
u
m=1 n=1 1
Then for σ > 0 we have
2
(4)
s
∞
ϕ(x)x−s−1 dx = (1 − 21−s )2 ζ 2 (s).
1
From (4) we obtain the following
Corollary 2.
∞
√
4
2 (5)
3 − 8 cos(t log 2) ζ( 21 + it) 0
dt
1
4
+
t2
2 = π
∞
ϕ2 (x)
1
dx
.
x2
The integral on the right-hand side of (5) is elementary, but nevertheless its evaluation in closed form is complicated.
Proof of Theorem 1. We start from (see e.g., [1, Chapter 1]) the identity
(1 − 21−s )ζ(s) =
(6)
∞
(−1)n−1 n−s
(σ > 0)
n=1
and
∞
(7)
−∞
cos(αx)
π
dx = e−|α|σ
σ 2 + x2
σ
(α ∈ R, σ > 0),
which follows by the residue theorem on integrating eiαz /(σ 2 + z 2 ) over the contour
consisting of [−R, R] and semicircle |z| = R, m z > 0 and letting R → ∞. By
using (6) and (7) it is seen that the left-hand side of (1) becomes
∞ it
∞
∞ m
dt
m+n
−σ
(−1)
(mn)
2 + t2
n
σ
−∞
m=1 n=1
∞
∞
cos(t log m
π
n)
= ζ(2σ) + 2
(−1)m+n (mn)−σ
dt
2
2
σ
σ +t
−∞
m=1 n<m
∞
m−1
m
π
ζ(2σ) + 2
(−1)m m−σ
(−1)n n−σ · e−σ log n
=
σ
m=1
n=1
∞
m−1
π
m −2σ
n
ζ(2σ) + 2
(−1) m
(−1)
=
σ
m=1
n=1
∞
π
π
2k
−2σ
ζ(2σ) + 2
(−1) (2k)
(−1) = (1 − 21−2σ )ζ(2σ).
=
σ
σ
k=1
22
Aleksandar Ivić
This holds initially for σ > 1, but by analytic continuation it holds for σ > 0 as
well.
We shall provide now a second proof of Theorem 1. As in the formulation of
Theorem 2, let χA (x) denote the characteristic function of the set A, and let the
interval [a, b) denote the set of numbers {x : a x < b}. Then, for σ > 0, we have
∞
x−s−1
1
(8)
∞
χ[2n−1,2n) (x) dx =
n=1
∞ n=1
2n
x−s−1 dx
2n−1
∞
(1 − 21−s ) ζ(s)
1 =
(2n − 1)−s − (2n)−s =
s n=1
s
in view of (6). Now we invoke Parseval’s identity for Mellin transforms (see e.g.,
[1] and [3]). We need this identity for the modified Mellin transforms, defined by
∞
∗
f (x) x−s−1 dx.
F (s) ≡ m[f (x)] :=
1
The properties of this transform were developed by the author in [2]. In particular,
we need Lemma 3 of [2] which says that
∞
1
f (x) g(x) x1−2σ dx =
F ∗ (s) G∗ (s) ds
(9)
2πi e s=σ
1
if F ∗ (s) = m[f (x)], G∗ (s) = m[g(x)], and f (x), g(x) are real-valued, continuous
functions for x > 1, such that
1
x 2 −σ f (x) ∈ L2 (1, ∞),
1
x 2 −σ g(x) ∈ L2 (1, ∞).
From (8) and (9) we obtain, for σ > 0,
2
∞
∞
(1 − 21−s )ζ(s) 2
1 1
1−2σ
ds.
χ
(x)
x
dx
=
[2n−1,2n)
x2 n=1
2πi e s=σ s
1
But as χ2A (x) = χA (x), it is easily found that the left-hand side of the above identity
equals
∞ ∞
∞ χ[2m−1,2m) (x)χ[2n−1,2n) (x)x−1−2σ dx
m=1 n=1 1
∞ 2n
=
n=1
2n−1
x−1−2σ dx =
(1 − 21−2σ )ζ(2σ)
2σ
in view of (6), and (1) follows.
For the Proof of Theorem 2 we need the following
23
Some identities for the Riemann zeta-function
Lemma. Let 0 < a < b. If f (x) is integrable on [a, b], then
2
b
−s
f (x)x
a
(10)
ab
=
−s
dx
x
a2
x du
x/a
f (u)f
u
a
u
b2
dx +
−s
b
x
f (u)f
x du
x/b
ab
u
u
dx.
The identity (10) remains valid if b = ∞, provided the integrals in question
converge, in which case the second integral on the right-hand side is to be omitted.
Proof. We write the left-hand side of (10) as the double integral
b
a
b
(xy)−s f (x)f (y) dx dy
a
and make the change of variables x = X/Y, y = Y . The Jacobian of this transformation equals 1/Y , hence the left-hand side of (10) becomes
b2
min(X/a,b)
X dY
−s
dX
X
f (Y )f
Y
Y
max(a,X/b)
a2
b2
ab
X/a
b
X dY
X dY
dX +
dX,
=
X −s
f (Y )f
X −s
f (Y )f
Y
Y
Y
Y
a2
a
X/b
ab
as asserted.
Proof of Theorem 2. We use (8) and the Lemma to obtain that (4) certainly
holds with ϕ(x) given by (3), since trivially ϕ(x) x. To see that it holds for
σ > 0, we note that
(11)
x
g(u)g
1
x du
u
u
=
√
x
+
1
and use (11) with
g(x) =
∞
x
√
x
=2
x
√
x
g(u) g
x du
,
u u
χ[2n−1,2n) (x).
n=1
Note then that the integrand in ϕ(x) equals 1/u for 2m − 1 u 2m, 2n − 1 u 2n, and otherwise it is zero. This gives the condition
√
√
4mn − 2m − 2n + 1 x < 4mn, 21 x n ≤ 21 (x + 1), 1 m 21 ( x + 1).
We also have
x
x
du
χ[2m−1,2m) ( )χ[2n−1,2n) (u)
√
u
u
x
2n
2n−1
du
1
.
u
2n − 1
24
Aleksandar Ivić
Therefore
ϕ(x) √
m x
x/(4m)<n(x−1+2m)/(4m−2)
1
n
m
x 1 + 2 log x.
m
√ x
(12)
m x
This bound shows that the integral in (4) is absolutely convergent for σ > 0. Thus
by the principle of analytic continuation this completes the proof of Theorem 2.
Corollary 2 follows then from (4) and (9) on setting σ = 21 .
It is interesting to note that the bound in (12) is actually of the correct order
of magnitude. Namely we have
Theorem 3. For any given ε > 0 we have
(13)
ϕ(x) =
1
4
log x +
1
2
log
π
2
1
+ Oε xε− 4 .
Proof. By (8) and the inversion formula for the Mellin transform m[f (x)] (see
[2, Lemma 1]) we have, for any c > 0,
(1 − 21−s )2 ζ 2 (s)xs
1
ds.
(14)
ϕ(x) =
2πi e s=c
s2
We shift the line of integration in (14) to c = ε − 1/4 with 0 < ε < 1/8, which
clearly may be assumed. Since ζ(0) = − 21 and ζ (0) = − 21 log(2π), the residue at
the double pole s = 0 is found to be
π
1
log x + A, A = −ζ (0) − log 2 = 21 log
.
(15)
4
2
We use the functional equation (see e.g., [1, Chapter 1]) for ζ(s), namely
χ(s) = 2s π s−1 sin( 21 πs)Γ(1 − s)
ζ(s) = χ(s)ζ(1 − s),
with
χ(s) =
Let s = ε −
1
4
2π
t
σ+it− 21
e
i(t+ 41 π)
· 1+O
1 t
(t 2).
+ it. Then by absolute convergence we have
(1 − 21−s )2 ζ 2 (s)xs
dt
s2
T
3 −2ε
2T
∞
1
(1 − 21−s )2 ε− 1 +it t 2
ε−5/4
4
=i
d(n)n
x
eiF (t,n) dt + O(T − 2 −2ε ),
2
s
2π
T
n=1
2T
Some identities for the Riemann zeta-function
25
where d(n) is the number of divisors of n and
d2
2
(t log x + F (t, n)) = − .
2
dt
t
F (t, n) := 2t + t log n − 2t log(t/2π),
Hence by the second derivative test (see [1, Lemma 2.2]) the above series is
∞
d(n)nε−5/4 T −2ε = ζ 2
5
4
− 2ε T −2ε T −2ε .
n=1
This shows that
e s=ε−1/4
(1 − 21−s )2 ζ 2 (s)xs
ds xε−1/4 ,
s2
hence (13) follows from (14), (15) and the residue theorem.
In concluding, note that if we write
ϕ(x) =
1
4
log x + A + ϕ1 (x),
where A is given by (15) then, for e s = σ > 0, (4) yields
∞
A
1
+ 2+
s2
ϕ1 (x)x−s−1 dx = (1 − 21−s )2 ζ 2 (s),
s
4s
1
and the above integral converges absolutely, for σ > −1/4, in view of (13). Thus
by analytic continuation it follows that, for σ > −1/4,
∞
2
1
As + 4 + s
ϕ1 (x)x−s−1 dx = (1 − 21−s )2 ζ 2 (s).
1
REFERENCES
1. A. Ivić The Riemann zeta-function. John Wiley & Sons, New York, 1985.
2. A. Ivić On some conjectures and results for the Riemann zeta-function and Hecke
series. Acta Arith. 109(2001), 115–145.
3. E.C. Titchmarsh Introduction to the Theory of Fourier Integrals. Oxford University
Press, Oxford, 1948.
Katedra Matematike RGF-a Universiteta u Beogradu,
11000 Beograd,
- ušina 7,
D
Serbia and Montenegro
E–mail: eivica@ubbg.etf.bg.ac.yu, aivic@rgf.bg.ac.yu
(Received February 5, 2003)